Aptitude - Calendar - Discussion
Discussion Forum : Calendar - General Questions (Q.No. 6)
6.
If 6th March, 2005 is Monday, what was the day of the week on 6th March, 2004?
Answer: Option
Explanation:
The year 2004 is a leap year. So, it has 2 odd days.
But, Feb 2004 not included because we are calculating from March 2004 to March 2005. So it has 1 odd day only.
The day on 6th March, 2005 will be 1 day beyond the day on 6th March, 2004.
Given that, 6th March, 2005 is Monday.
6th March, 2004 is Sunday (1 day before to 6th March, 2005).
Discussion:
114 comments Page 5 of 12.
Roji said:
1 decade ago
Hey guys in this problem this formula not working.
day+mon code+(x-1990)+(x-1990)/4.
Can anyone explain in this type method.
day+mon code+(x-1990)+(x-1990)/4.
Can anyone explain in this type method.
Abinash Bhoi said:
1 decade ago
It is so easy.
One year less from March 2004 to March 2005. It is not including Fab 2004. So it is one year mean 1 odd day. 6 March 2005 is Monday so 6 March 2004 will be Sunday.
One year less from March 2004 to March 2005. It is not including Fab 2004. So it is one year mean 1 odd day. 6 March 2005 is Monday so 6 March 2004 will be Sunday.
Bama Karmakar said:
1 decade ago
As cover a year i.e 365 days.
365/7 52 week and 1 day remaining.
So we can say before 1 day from monday i.e Sunday.
365/7 52 week and 1 day remaining.
So we can say before 1 day from monday i.e Sunday.
Maukuti said:
1 decade ago
The answer provided here is wrong. Correct answer is SATURDAY as 2004 is a leap year. It has 2 odd days. So each date will be forward by 2 weekdays.
Anshu said:
1 decade ago
Why didn't we add odd days of 2005.
31+28+6/7=2 odd days.
And there is 1 odd day of 2004.
So there are 3 odd days.
What to do ?
31+28+6/7=2 odd days.
And there is 1 odd day of 2004.
So there are 3 odd days.
What to do ?
Yash Prajapati said:
1 decade ago
Yes this answer is absolutely correct ! because if there is given questions like this, in which it is given difference of 365 days that is 1 year there is always a difference of 1 day but if there is a difference of 366 days that is a leap year it becomes a difference of 2 days.
Sumit said:
1 decade ago
Anybody please tell me how to find leap year?
Anees said:
1 decade ago
Any given year is divisible by the number 4 is a leap year, pretty simple.
For e.g 2008/4 = 502 (NO REMAINDER).
2007/4 = 501.75 (REMAINDER).
Just do observe it's too simple.
For e.g 2008/4 = 502 (NO REMAINDER).
2007/4 = 501.75 (REMAINDER).
Just do observe it's too simple.
Sumanth geras said:
1 decade ago
Hi @Roji I can explain by your formula.
Day+Month code+(x-1990)+(x-1990)/4.
Step 1: First you should calculate the day for the first mentioned date.
i.e 4 for 6th march 2005.
[6+4+105+26=141/7 it should be Sunday since rem is 1].
But it was given Monday so it is shifted by one day.
Step 2: If you calculate for 6th mar' 2004 you will get Saturday.
So to get answer you should shift the obtained answer by one day. It works for all problems.
Day+Month code+(x-1990)+(x-1990)/4.
Step 1: First you should calculate the day for the first mentioned date.
i.e 4 for 6th march 2005.
[6+4+105+26=141/7 it should be Sunday since rem is 1].
But it was given Monday so it is shifted by one day.
Step 2: If you calculate for 6th mar' 2004 you will get Saturday.
So to get answer you should shift the obtained answer by one day. It works for all problems.
Manish said:
1 decade ago
Answer is wrong. Its Saturday. Check in calendar. 2004 is leap year so odd days should be 2 not 1.
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